解:(I)∵an=2an-1+2n
∴an2n=an-12n-1+1
即an2n-an-12n-1=1
∴数列{an2n}是等差数列,公差为=1,首项为a12=12
∴an2n=12+(n-1)×1
∴an=(2n-1)•2n-1
(II)Sn=1•20+3•21+5•22+…+(2n-1)•2n-1
∴2Sn=1•21+3•22+…+(2n-3)•2n-1+(2n-1)•2n
两式相减得
-Sn=1+2•21+2•22+…+2•2n-1-(2n-1)2n=(3-2n)•2n-3
∴Sn=(2n-3)•2n+3